Let $a\in R$ for any any $\epsilon > 0$ we know that $(a-\epsilon,a+\epsilon)$ contains both rational and irrational points. Therefore, all points are boundary points. Thus, $\partial \mathbb{Q} = \mathbb{R}$ which means $\text{cl}( \mathbb{Q}) = \mathbb{Q} \cup \partial \mathbb{Q} = \mathbb{R}$.